Question: $ g(x) = \int_{1}^x\sqrt{19 - t}\,dt\,$ $ g'(3)\, =$
Solution: The Fundamental Theorem of Calculus If $~ g(x)=\int_a^xf(t)\,dt\,$, then $~g^\prime (x)=f(x)\,$ This only works if $f$ is continuous on $[a,b]$. Thankfully, the function $f(t) = \sqrt{19-t}$ is continuous on $[1,3]$. Applying the theorem We're given: $ g(x) = \int_{1}^x\sqrt{19 - t}\,dt\,$ So the theorem tells us: $ g\,^\prime(x) =\sqrt{19 - x}$ Evaluating $g'(3)$ $ g'(3) = \sqrt{19 - 3}=\sqrt{16}=4$ The answer: $g'(3)=4$